please give the code to java
please give the code to java

Edited by author 18.11.2017 07:25

#include<iostream>
#include<cstdio>
using namespace std;

int main()
{

        int n, a = 1, counter = 0;
        string s;
        cin >> n;

        for (int i = 0; i < n; ++i)
         {
          cin >> s;
                if (s[0] == 'A' || s[0] == 'P' ||s[0] == 'O' ||s[0] == 'R' &&  a == 2){ a = 1; counter++;}
            else
                 if (s[0] == 'A' || s[0] == 'P' ||s[0] == 'O' ||s[0] == 'R' &&  a == 3){ a = 1; counter +=2;}

                 if (s[0] == 'B' || s[0] == 'M' ||s[0] == 'S' &&  a == 1){ a = 2; counter++;}

            else

                 if (s[0] == 'B' || s[0] == 'M' ||s[0] == 'S' &&  a == 3){ a = 2; counter++;}


                if (s[0] == 'D' || s[0] == 'G' ||s[0] == 'K' ||s[0] == 'T' || s[0] == 'W' &&  a == 1){ a = 3; counter +=2;}

            else

                if (s[0] == 'D' || s[0] == 'G' ||s[0] == 'K' ||s[0] == 'T' || s[0] == 'W' &&  a == 2){ a = 3; counter++;}



        }

        cout << counter;
        return 0;

}

initially  let result sequence to be 10*  10* 10*   0* means zero or more zeroes.

every time choose two number and if we merge them produce a smaller number then we merge them.

the procedure stops until we can't merge to get a smaller number

is it correct?? seems to be..

Edited by author 16.11.2017 07:10

Anyone crash test#2

aaaaa
aaa

No
aaa

aaaaa
aaaaaaaaa

No
aaaa aaaaa

aabaa
aabaabaabaaaaabaa

No
aab aab aaba aa aabaa

aabaa
aabaabaab

No
aab aab aab

abcabcabccs
abcaabcaabcaa

No
abca abca abca a

aabaa
aabaabaaaa

No
aab aabaa aa

What's wrong? I have 'time limit excedeed' in 9th test.



import functools

def pr(n):
    result = []
    i = 2
    now = 1
    while n >= 1:
        if n % i == 0:
            now *= i
            n //= i
        elif n % i != 0 and now > 1:
            result.append(now)
            i += 1
            now = 1
        else:
            i += 1
        if n == 1:
            result.append(now)
            return result

def f(n):
    result = 0
    i = 1
    while i ** 2 < n:
        if n % i == 0:
            result += 1
        i += 1
    if i ** 2 == n:
        return 2 * result + 1
    else:
        return 2 * result

newList = [int(input()) for _ in range(10)]
product = functools.reduce((lambda x, y: x * y), newList)
print(functools.reduce(lambda x, y: x * y, map(f, pr(product))) % 10)

You don't must develop hard solution. Your work should not payed many time. Just pay attention, and look again on tests. Attention will save your time.
I am sorry, for my poor English. Good luck!

Edited by author 18.10.2015 16:52

Edited by author 11.11.2017 22:53

Edited by author 11.11.2017 22:53

I was trying to write TRIE for mnemonic words and DFS that tree with phone number.
And I have failed.
After that,  I wrote just 2-D dimensional dynamic programming solution.  DP[i] [j]  is minimal number of words to memorize substring [i;j].  And I got success.  Used Python.  Just input,  not stdin from sys.

using System;
namespace t1068{
    class Program{
        public static void Main(string[] args){
            int n = int.Parse(Console.ReadLine());
            int i = 0;
            while (n != 1){
                if (n >= 1){
                    i+= n;
                    n--;
                }
                else if (n <= 0){
                    i+= n;
                    n++;
                }
            }
            Console.Write(i + 1);
            //Console.ReadKey();
        }
    }
}

time 0.078
memory 3 350 kb.

I use triplex tree.
It is very interesting problem.

#include <stdio.h>
#include <iostream>
using namespace std;
int n, m, segRez[3004];
int segm1[3004];
char segm2[6005];
int main()
{
    cin >> n >> m;
    for(int i = 0; i < n; ++i){
        gets_s(segm2);
        for(int j = 0; j < m; ++j){
            segRez[j] = segm1[j] + int(segm2[j<<1] - 48);
            segm1[j] = int(segm2[j<<1] - 48);
            if(j - 1 >= 0 && ((segRez[j - 1]|segRez[j])) == 3){
                cout << "No";
                return 0;
            }
        }
    }
    cout << "Yes";
}

Here is my java code :


import java.util.Scanner;
public class Example{
    public static void main(String []args){
        Scanner in = new Scanner(System.in);
        int number1,number2,number3;
        int [] array;
        int position=0;
        boolean check=false;
        int sth;
        number1=in.nextInt();
        number2=in.nextInt();
        number3=in.nextInt();
        array = new int[number2];
        for(int i=0;i<=number2-1;i++){
            sth=(int)Math.pow(i,number1);
            if((sth-number3)%number2==0){
                array[position]=i;
                position++;
                check=true;
            }else if(i==number2-1 && check==false){
                System.out.println(-1);
                System.exit(0);
            }

        }
        for(int p=0;p<position-1;p++){
            for(int j=p+1;j<position;j++){
                if(array[p]>array[j]){
                    int t=array[p];
                    array[p]=array[j];
                    array[j]=t;
                }
            }
        }
        for(int i=0;i<position;i++){
            System.out.print(array[i] + " ");
        }
        System.out.println();
    }
}

I get an error on test#3 .
And on my C++ solution i get an error on test#6.



#include<iostream>
#include<cmath>
using namespace std;
int main(void){
unsigned short int N,M,Y;
cin>>N>>M>>Y;
unsigned short int X[M];
unsigned short int position=0;
bool some_check=false;
int sth;
for(int i=0;i<=M-1;i++){
    sth=pow(i,N);
    if((sth-Y)%M==0){
        X[position]=i;
        position++;
        some_check=true;
    }
    else if(i==M-1 && some_check==false){
        cout<<-1<<endl;
        return 0;
    }
}
for(unsigned short int i=0;i<position-1;i++){
    for(unsigned short int j=i+1;j<position;j++){
        if(X[i]>X[j]){
            swap(X[i],X[j]);
        }
    }
}
for(unsigned short int i=0;i<position;i++){
    cout<<X[i]<<" ";
}
cout<<endl;
return 0;
}


Please,any ideas ???

int main()
{
    int n;
    string s1, s2;
    cin >> n >> s1 >> s2;
    s1 += s1;
    int pos = (int) s1.find(s2);
    switch ( pos )
    {
    case -1:
        cout << -1;
        break;
    case 0:
        cout << 0;
        break;
    default:
        cout << n-pos;
    }
    return 0;
}

¿Any idea of why a code can pass all the tests except the test 6 ? I prove my code with this imput, and it works fine

7325189087
5
it
your
reality
real
our
4294967296
5
it
your
reality
real
our
213456
12
id
a
h
lm
aid
aidhlm
hl
m
dhlm
ai
aid
hlm
2017697240238123468420176972402381234684201769724023812346842017697240238123468420176972402381234684
7
azipmwpbgqadtjafhmth
azipmwpbgqadtjafhmth
azipmwpbgqadtjafhmth
azipmwpbgqadtjafhmth
azipmwpbgqadtja
azipmwpbgqadtjafhmthazipmwpbgqadtjafhmthazipmwpbgqadtjafhmthazipmwpbgqadtjafhmth
fhmth
27386428376
0
97854668689678768
3
dfgdfhjghjdggzash
dfgkjt
nkhhdfsge
11111111111
5
ij
jj
ji
j
i
1234567890
5
bdh
txo
bdhkmp
i
kmp
-1

This is my code:

import time
import sys
sys.setrecursionlimit(100000)

n = int(input("Enter N (for digits): "))
k = int(input("Enter K: "))

rangeVar = pow(k, n) - pow(k, n-1)
starter = pow(k, n) - rangeVar
counter = 0

start = time.clock()
def calc(i):
    global starter, rangeVar, counter, n, k

    temp = str(i)

    if i == pow(k, n):
        return counter
    else:
        for j in range(len(temp) - 1):
            if temp[j] == '0' and temp[j+1] == '0':
                counter += 1
        calc(i+1)

calc(starter)
end = time.clock()
print(rangeVar - counter)
print(end - start)

Remove repetitions before applying your algo i.e.: sequences like 2 1 2 2 2 2 1 change into 2 1 2 1
It helped me with WA#21